Singularities at Infinity implies constant or polynomial function - Proof? I have been doing some leisurely reading on complex analysis, and have came across several texts that mention the following:

*

*If an entire function has a removable singularity at infinity, then it must be constant

*If an entire function has a pole of order $k$ at infinity, then it must be a polynomial of degree at most $k$
I think this is a fascinating result - but I am struggling to figure out why exactly these are true.
I of course know that if $f$ has a removable singularity at infinity, then $g(z) = f(1/z)$ has a removable singularity at $0$. I have been playing around trying to get somewhere I can apply Liouville's Theorem, but can't quite get there.
For the second point, I know that if $f$ has a pole of order $k$ at infinity, then $g(z)=f(1/z)$ has a pole of order m at 0. This means there is some entire function $h$ where $g(z) = z^kh(z)$ and $h(0)\neq 0$. I was looking to bound this by a polynomial (as this would give the result), but not sure how to do this either...
Anyone able to help or provide proof would be hugely appreciated - keen to understand this better before reading further. :)
Thanks!
 A: Welcome.
If $f:\Bbb C\setminus B_r(0)\to\Bbb C$ is holomorphic (some $r>0$) then it can be shown that there exist constants $(\alpha_n)_{n\in\Bbb Z}\subset\Bbb C$ such that: $$f(z)=\sum_{n=-\infty}^\infty\alpha_nz^{-n}$$For all $z,|z|>r$ (the series expansion "at infinity").
If we are talking about an entire function $f$, then $r$ may be anything $>0$. In particular, the above series expansion (for the same constants) can be shown to be valid for every single $z\neq0$. Importantly, $f$ is still continuous (holomorphic, even) at zero, so $\lim_{z\to0}f(z)=f(0)$ must exist. Considering the above series expansion, it must be that $\alpha_n=0$ for all $n>0$ (otherwise, we get $0^{-n}\cdots$ problems).
Your second bullet point is just the first, with $k=0$. If $f$ has a pole of order $k$ at infinity, that means to say that $\alpha_n=0$ for all $n<-k$ in the above series and $\alpha_{-k}\neq0$. Tying that together with $\alpha_n=0$ for $n>0$, we get: $$f(z)=\sum_{n=-k}^0\alpha_nz^{-n}=\sum_{n=0}^k c_nz^n$$For some constants $c_n:=\alpha_{-n}$. Since $c_k$ is nonzero, by hypothesis on the order of the pole, that means $f$ is a polynomial of order $k$ (with coefficients $c_n$ as shown).

Explicitly, we determine the Laurent coefficients $\alpha_n$ as follows. Let $\gamma$ be any simple, closed contour around zero that is contained in $B_{1/r}(0)$. Then: $$\alpha_n=\oint_\gamma\frac{f(1/w)}{w^{n+1}}\,\mathrm{d}w$$The derivation is basically the same as the usual Laurent series derivation, and you can even apply the usual derivation to $g:z\mapsto f(1/z)$. Notice that if $f(z)$ as a limit as $z\to0$, $f(1/w)$ has a definite limit as $w\to\infty$. Moreover, $r>0$ can be anything (in the entire case) so $B_{1/r}(0)$ can be expanded to reach any particular point in $\Bbb C$. So, we can let $\gamma$ be the circle of radius $R$ for any $R>0$ (large) (without changing the value of $\alpha_n$) to see that: $$|\alpha_n|\le\oint_{C_R}|f(1/w)||w|^{-n-1}\,|\mathrm{d}w|\le R^{-n-1}\cdot\max_{|w|=R}|f(1/w)|\cdot2\pi R=\mathcal{O}(R^{-n})\to0,\,n>0$$Since $|f(1/w)|\to|f(0)|$. Hence, $\alpha_1,\alpha_2,\cdots$ are all equal to zero.
Notice this is essentially the same as the proof of Liouville's theorem.
A: For your first point, take $f$ to be entire with a removable singularity at infinity.
Then $g(z)=f(1/z)$ has a removable singularity at $0$ like you have said.
If we examine the power series for $f$, which is $f(z)=\sum_{n\geq 0} c_nz^n$ then we can consider
$$g(z) = \sum_{n \geq 0} c_n z^{-n}$$
So as $|z| \to \infty$ we have $|g(z)| \to c_0$.
Thus $g$ is bounded on $\{ z : 0 < |z| < r\}$
Consequently, $f$ is bounded on $\{ z : |z| >\frac{1}{r}\}$.
If we consider $\{z : |z| \leq \frac{1}{r}\}$ then since this is compact, and $f$ holomorphic (and so continuous), $f$ must be bounded on this set also.
Therefore $f$ is bounded on all of the complex plane, so Liouville’s Thm dictates it must be constant.
